Question

One die has three faces marked with $1$; two faces marked with $2$ , and one face marked with $3$; another has one face marked$1$ , two marked $2$ and three marked $3$, then
A. The most probable throw with two dice is $4$
B. The probability of most probable throw is $\dfrac{1}{4}$
C. The probability of most probable throw is$\dfrac{7}{18}$
D. None of these

Answer 1

Hint: In statistics, probability tells us how often some event will happen after many repeated trials. The probability of an event is a number between $0$ and $1$, where, roughly speaking, $0$ indicates impossibility of the event and $1$ indicates certainty.

Complete step by step solution:
Let us take dice A and B having six faces each. The events occurrence when we throw a dice A are $\left\{ 1,1,1,2,2,3 \right\}$. The events occurrence when we throw a dice B are $\left\{ 1,2,2,3,3,3 \right\}$. The total occurrence of events is $6$. Now, we have to find the probability of $1$, $2$, and $3$ .
We know the probability formula which is:
$\Rightarrow P\left( A \right)=\dfrac{n\left( A \right)}{N}$ where $n\left( A \right)$ is the number of outcomes and $N$ is the total number of outcomes.
 Now we will find the probability of dice A.
The probability of getting $1$ or $p\left( 1 \right)$ is equal to $\dfrac{3}{6}=\dfrac{1}{2}$ and probability of getting $2$ or $p\left( 2 \right)$ is equals to $\dfrac{2}{6}=\dfrac{1}{3}$ and the probability of getting $3$ or $p\left( 3 \right)$ is equals to $\dfrac{1}{6}$ .
Similarly, the probabilities of dice B are:
$\Rightarrow p\left( 1 \right)=\dfrac{1}{6}$
$\Rightarrow p\left( 2 \right)=\dfrac{1}{3}$
$\Rightarrow p\left( 3 \right)=\dfrac{1}{2}$
The sum of occurrence of the dice A and B are:
$\Rightarrow sum=\left\{ 2,3,4,5,6 \right\}$
Now
$\Rightarrow sum2=1+1$
The probability of $p\left( 2 \right)$ is
$\Rightarrow p\left( 2 \right)=\left( \dfrac{1}{2} \right)\left( \dfrac{1}{6} \right)=\dfrac{1}{12}$
Again sum of $3=1+2$
The probability of $p\left( 3 \right)$ is
$\begin{align}
  & \Rightarrow p\left( 3 \right)=\left( \dfrac{1}{2} \right)\left( \dfrac{1}{3} \right)+\left( \dfrac{1}{3} \right)\left( \dfrac{1}{6} \right) \\
 & \Rightarrow p\left( 3 \right)=\dfrac{4}{18} \\
\end{align}$
Again sum of $4=1+3,2+2,3+1$ then,
$\begin{align}
  & \Rightarrow p\left( 4 \right)=\left( \dfrac{1}{2} \right)\left( \dfrac{1}{2} \right)+\left( \dfrac{1}{3} \right)\left( \dfrac{1}{3} \right)+\left( \dfrac{1}{6} \right)\left( \dfrac{1}{6} \right) \\
 & \Rightarrow p\left( 4 \right)=\dfrac{7}{18} \\
\end{align}$
Again sum of $5=2+3,3+2$, then
$\begin{align}
  & \Rightarrow p\left( 5 \right)=\left( \dfrac{1}{3} \right)\left( \dfrac{1}{2} \right)+\left( \dfrac{1}{6} \right)\left( \dfrac{1}{3} \right) \\
 & \Rightarrow p\left( 5 \right)=\dfrac{2}{9} \\
\end{align}$
And the last is sum of $6=3+3$ then
$\begin{align}
  & \Rightarrow p\left( 6 \right)=\left( \dfrac{1}{6} \right)\left( \dfrac{1}{2} \right) \\
 & \Rightarrow p\left( 6 \right)=\dfrac{1}{12} \\
\end{align}$
Therefore we get from these above data $\dfrac{7}{18}$ is the maximum probability of throwing.
Hence $\dfrac{7}{18}$ is the maximum probability of throwing for the sum of $4$.
So, the correct answer is “Option C”.

Note: We can go wrong by taking the sum of the occurrence, here we take the sum of occurrence of dice A and dice B because we need the maximum occurrence of both dies. Sometimes we made a mistake here not to take the sum.